"Unexpected identifier" Error in lua trigger code

Plugins & Plugin Development Alternate UI
1

HI,
Just decided to implement a new TV control in my Vera, and thought I would try it in Workflow.
I have created the states and transitions, but is not detecting my TV coming on.
I am monitoring a device status, with
new==‘1’
in the Lua expression.

but this is giving an error - saying “unexpected identifier new near 1” (see screenshot)
Any sort of entry seems to give a similar problem… any idea on what I am doing wrong gratefully received.

Thanks

2

That is an illegal expression (as it is telling you)

This is a a comparison operator and should be used as part of a conditional expression or assignment.

i.e.
if ( new == ‘1’) then
print(“New has a value of 1”)
end

or
TestNew = new == ‘1’

In the latter case TestNew will have a value of true or false.

3

Thanks for prompt response Richard, but this is not a standalone piece of Lua code, it is a condition that sits within a evaluation for a trigger in the State machine.
ie I believe this is the
if ( [LUA EXPRESSION]) then
[EXECUTE TRIGGER CHANGE OF STATE]
print(“New has a value of 1”)
end
part of your example.

In Report format my code looks the same of other examples in the ALtui thread.

4

You will need to post the WHOLE expression and not a fragment.
Syntax errors else where can cause it to complain here … i.e. a missing single quote previously.

5

Richard,
new ==‘1’
is the whole expression!
It appears that the editor is correctly flagging a syntax error, but that is a red herring in this context as that snippet is all it needs, or at least, the state machine is working despite the warning.

False alarm.

Thanks for your response

6

See this post for info on the errant warning: http://forum.micasaverde.com/index.php/topic,48034.0.html

It’s a trade off between getting formatting versus the red X.